Alcohols, Phenols and Ethers - 1 File Download...Alcohols, Phenols and Ethers Solutions Chapter 11...
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Level - II
SECTION - A
Objective Type Questions
(Preparation of Alcohols and Phenols)
1. CH CH–CH=CH3 2
CH3
(i) B H2 6 H SO2 4
(ii) H O /OH2 2
–
140°X Y
What is Y?
(1) CH –CH–CH –CH –O–CH –CH –CH–CH3 2 2 2 2 3
CH3
CH3
(2) CH –CH–CH=CH3 2
CH3
(3) CH –CH–CH–O–CH–CH–CH3 3
CH3CH
3CH
3CH
3
(4) CH –C–O–C–CH3 3
C H2 5
CH3
CH3
C H2 5
Sol. Answer (1)
(1) B H 2 6
(2) H O /OH 2 2
– OH
H SO , 140°2 4
O
Alcohols, Phenols and Ethers
Solutions
Chapter 11
122 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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2. + H SO (fuming)2 4
X YNaOH
570-620K
What is Y?
(1)
OH
(2)
SO H3
(3)
ONa
(4)
Sol. Answer (3)
+ H SO2 4
(Fuming)
SO H3
NaOH
570-620 K
O Na+–
Y
3. CH2
CH CH2
Br(1) Mg
(2) HCHO
(3) H O3
+
ABr
2
CCl4
B
KOH
C
Product (C) is
(1)O
(2)O..
..
(3)
CH2
CH CH2
CH2
OH
OH OH(4)
O
HO
HO
Sol. Answer (3)
+ MgBr MgBr
HCHO, H+
CH OH2
(1) H , Br , CCl KOH+
2 4 4,
OH
OHOH
4. Ethylene oxide when treated with Grignard reagent yields
(1) Primary alcohol (2) Secondary alcohol (3) Tertiary alcohol (4) Cyclopropyl alcohol
Sol. Answer (1)
O+ RMgX R OH
(1° alcohol)
123Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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5. Cl Br A BMg/ dry ether (I) CH –C–CH
3 3
(ii) H O3+
O
(1 mole)
What is B?
(1) Cl C–CH3
OH
CH3
(2) CH –C3
C–CH3
OHOH
CH3CH
3
(3) CH –C3 Br
OH
CH3
(4)
Sol. Answer (1)
∵ Br has more tendency to form RMgX, than 'Cl' because of more size. So, reaction will take place from
'Br' side. i.e.,
Mg/dry ether
(1 mole)BrCl MgBrCl CCl
O–
Mg Br+
CH3
CH3
CCl
OH
CH3
CH3
(ii). H O 3
+
C
CH3
(i). H C3
O
(A)
(B)
6. + HNO3 X YH SO
2 4 Sn + HCl
Z ANaNO + HCl
2H O
2
0-5°C. What is A?
(1)
NH2
(2)
NO2
(3) (4)
OH
Sol. Answer (4)
+ HNO3
NO2
Sn/HCl
NH2
NaNO ,dil
HCl2
H SO2 4
(X)
N Cl2
+ –
H O2
OH
(A)
124 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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(Physical and Chemical properties of Alcohols and Phenols)
7. + H SO2 4
A
CH OH2
(conc.)170°C
What is the major product A?
(1)
CH2
(2)
CH3
(3) (4)
CH3
Sol. Answer (3)
+ H+
H
(Cyclohexene)
OH
8.H
+ CH –CH –CH –OH3 2 2
ABr /Fe
2
B
What is the major product B?
(1)CH –CH –CH
32 2
(2) CH –CH–CH23
Br Br
(3)
CH –3C–Br
CH3
(4)
CH –3CH
CH3
Br
Sol. Answer (4)
OH + H
+
+
1 – 2 H:
Shift
–
+
Br /Fe2
Br
125Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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9. Which of the following is the correct increasing order of boiling point of following compounds?
CH –CH –CH –OH23 2
CH –CH–CH3 3
CH –O–CH –CH3 2 3
I II III
OH
(1) II < I < III (2) III < II < I (3) I < II < III (4) II < III < I
Sol. Answer (2)
The order of boiling point is
OH>
(I)
> OOH
(II) (III)
10. Which of the following is the correct ease of dehydration?
CH –CH –CH –CH3 2 2 2
CH –CH –CH–CH3 2 3
I II
OH
OH OH
CH =CH–CH–CH2 3
III IV
OH
(1) I > III > II > IV (2) IV > III > II > I (3) IV > II > III > I (4) III > IV > II > I
Sol. Answer (2)
The ease of dehydration is,
OH
>
OH
>
OH
> OH
(I)(II)(III)(IV)
11. CH – C = CH3 2
CH3
H O3
(1) B H /THF2 6
(2) H O /OH2 2
A
B
Product A and B can be distinguished by
(1) Sodium metal
(2) Neutral FeCl3
(3) Lucas reagent
(4) Esterification reaction
Sol. Answer (3)
A
B
(3°)
(1°)
OH
OH
Lucas reagent can distinguish them
126 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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12. CH – C – OH3
CH3
CH3
Al O /2 3
A B
HBr
CH –O–O–CH3 3
aq. NaOHCD
MgBr
The end product D of the reaction is
(1)O – C(CH
3)3
(2)(CH ) – C – CH
3 3 2
(3) (4)
OH
BrBr
Br
Sol. Answer (3)
A
C
OHB
D
OH
Br
13. Consider the following reaction :
3
3
CH Cl
Zn dust Anhydrous AlClPhenol X Y
4
2
Alkaline KMnO
H O/HZ
The product Z is
(1) Benzene (2) Toluene
(3) Benzaldehyde (4) Benzoic acid
Sol. Answer (4)
OH
Zn CH Cl3
AlCl3
CH3
(Y)(X)
Alkaline
KMNO4
COOH
(Z)
14. CH –CH–CH + HI2 2 X
OH OH OH
(excess)
What is X?
(1) CH –CH–CH22
I I I
(2) CH =CH–CH2 2
I
(3) CH3–CH=CH
2(4) CH –CH–CH
33
I
127Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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Sol. Answer (4)
H C2
CH CH + HI2
(excess)
OH OH OH
H C3
CH CH3
I
HI
H C2
CH CH2
I I I
–I2
H C2
CH CH2
HIH C
2CH CH
3
I I
H C3
CH CH3
I
HIH C
2CH CH
3
–I2
I
Answer (4)
15. + HIred P
(1 eq.)
XBr
2
Y
OH
CH OH2
h
What is Y?
(1)
I
CH Br2
(2)
OH
CH I2
Br
(3)
OH
CH Br2
(4)
CH Br2
Sol. Answer (3)
OH
OH
+ HI
OH
CH3
Br2
h
OH
Br
Red P
16. Which of the following will not give positive test with neutral FeCl3?
(1) Nitrophenol (2) Phenol (3) Allyl alcohol (4) o-cresol
Sol. Answer (3)
+ FeCl3
OHNo Reaction
17. In Dow's process haloarene is converted to phenol with fused NaOH. The most reactive compound is
(1)
Cl
H
H
H
H
H
(2)
Cl
D
D
D
D
D
(3)
Cl
T
T
T
T
T
(4) All are equally reactive
128 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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Sol. Answer (1)
Cl
H
H
H
H
H
18. Among the following four compounds
a. Phenol b. Methyl phenol c. Meta nitrophenol d. Para nitrophenol
The acidity order is
(1) d > c > a > b (2) c > d > a > b (3) a > d > c > b (4) b > a > c > d
Sol. Answer (1)
The acidic order is
Paranitrophenol > Meta-nitrophenol > Phenol > Methyl Phenol
19.
C OH
H
CH3
SOCl2
PyridineProduct (major)
The product is
(1)
C H
Cl
CH3
(2)
C Cl
H
CH3
(3) Mixture of (1) & (2) (4) No reaction
Sol. Answer (1)
In absence of pyridene, the reaction of SOCl2
with alkyl halide is SNi (Intramolecular nucleophilic substitution
reaction) and hence SNi, the confuguration is retained.
i.e., retention reaction takes place in SNi. So,
H C3
OH
SOCl2
H
C C
H C3
H
Cl
20.
CH2
HCl
1 eqv.A
aq KOHB
conc. H SO42
C (major). Product (C) is
(1)
CH3
(2)
CH2
(3) (4)
129Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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Sol. Answer (2)
CH2
HCl
leq.
CH2 – Cl
(A) (B)
aq.KOH
CH2 – OH
Conc. H SO2 4
CH2
(C)
.
(Ethers-Preparation and properties)
21. Reaction of t-butyl bromide with sodium methoxide produces
(1) Sodium t-butoxide (2) t-butyl methyl ether (3) Isobutane (4) Isobutylene
Sol. Answer (4)
Br + CH O Na3
– +
(Isobutylene)
22. C CH2
OCH
3
CH3
CH MgBr
H O
3
3
+
Cu
300ºCA B
B is
(1) (CH3)3CHO (2) CH – C = CH – CH
3 3
CH3
(3) (CH3)2CHCOCH
3(4) CH C – CH – CH CH
3 2 2 3
O
Sol. Answer (2)
OCH MgBr
3
H O3
+
Cu, 300°C
(B)
OH
In basic medium the opening of epoxide is SN2 type. So, nucleophile CH
3 attack less hindered 'C' of
epoxide.
23. Ethyl chloride is converted into diethyl ether by
(1) Perkin’s reaction (2) Grignard reaction
(3) Wurtz synthesis (4) Williamson’s synthesis
Sol. Answer (4)
C H Cl2 5 o
This can be done by Williamson's synthesis.
24.O CH OH
2
PCl5
AMg
ether B
CH3
CH CH2
OC
H O3
+
Product (C) is
(1)O O
(2)O
OH
(3)
O
OH
(4)
O
OH
130 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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Sol. Answer (3)
O CH OH2
PCl5
O Cl
Mg
O MgCl
o
OOH
(A)
(B)
(C)
25. A B
O
HI/ alc. KOH/excess C
Cl /h2
Product C is
(1) Alkyl iodide (2) Vinyl chloride (3) Vinyl iodide (4) Allyl chloride
Sol. Answer (4)
Alc. KOH/ HI / Cl /h2
(A) (B) (Allyl chloride)
Cl
I I
SECTION - B
Previous Years Questions
1. The reaction that does not give benzoic acid as the major product is [NEET-2019 (Odisha)]
(1)
CH OH2
KMnO /H4
+
(2)
CH OH2
K Cr O2 2 7
(3)
COCH3
(i) NaOCl
(ii) H O3
+(4)
CH OH2
PCC
Pyridiniumchlorochromate⎛ ⎞⎜ ⎟⎝ ⎠
Sol. Answer (4)
CH2
PCC
Pyridiniumchlorochromate⎛ ⎞⎜ ⎟⎝ ⎠
OHCHO
PCC oxidises primary alcohol to aldehyde.
2. The major products C and D formed in the following reaction respectively are
excessHI
3 2 2 3 3H C–CH –CH –O–C CH C D [NEET-2019 (Odisha)]
(1) H3C – CH
2 – CH
2 – OH and HO – C(CH
3)3
(2) H3C – CH
2 – CH
2 – I and I – C(CH
3)3
(3) H3C – CH
2 – CH
2 – OH and I – C(CH
3)3
(4) H3C – CH
2 – CH
2 – I and HO – C(CH
3)3
131Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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Sol. Answer (2)
H C – C – C – O – C(CH )3 2 2 3 3
H HHI
C – C – C – OH + (CH )H H H C – I3 2 2 3 3
HI
C C C –I H H H3 2 2
3. The structure of intermediate A in the following reaction, is [NEET-2019]
CHCH
3
CH3
O2
AH
+
H O2
OH
+ H C3
O
CH3
(1)
O – CH
CH3
CH3
(2)
H C – C – O – O – H3
CH3
(3)
O – O – CH
CH3
CH3
(4)
HC
CH – O – O – H2
CH3
Sol. Answer (2)
CHCH
3
CH3
O2 H+
H O2
OH
+ H C – C –CH3 3
O
H C – C – O – O – H3
CH3
(A)Cumene
hydroperoxide
Cumene
4. The compound that is most difficult to protonate is :
[NEET-2019]
(1) H H
O
(2) H C3 H
O
(3) CH3
O
H C3 (4) Ph H
O
Sol. Answer (4)
Due to involvement of lone pair of electrons in resonance in phenol, it will have positive charge (partial), hence
incoming proton will not be able to attack easily.
5. The compound A on treatment with Na gives B, and with PCl5 gives C. B and C react together to give diethyl
ether. A, B and C are in the order [NEET-2018]
(1) C2H
5OH, C
2H
6, C
2H
5Cl (2) C
2H
5OH, C
2H
5Cl, C
2H
5ONa
(3) C2H
5OH, C
2H
5ONa, C
2H
5Cl (4) C
2H
5Cl, C
2H
6, C
2H
5OH
132 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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Sol. Answer (3)
C H OH2 5
NaC H O Na
2 5
– +
(A)
PCl5
C H Cl2 5
(C)
C H O Na2 5
– +
+ C H Cl2 5
S 2NC H OC H
2 5 2 5
(B)
(B) (C)
6. In the reaction [NEET-2018]
OH O Na– +
CHO
+ CHCl + NaOH3
The electrophile involved is
(1) Dichloromethyl cation 2
CHCl
(2) Formyl cation CHO
(3) Dichlorocarbene 2: CCl (4) Dichloromethyl anion 2
CHCl
Sol. Answer (3)
It is Reimer-Tiemann reaction. The electrophile formed is :CCl2 (Dichlorocarbene) according to the following reaction
–
3 3 2CHCl OH CCl H O ���⇀
↽���
..
..
–3 2
Electrophile
CCl : CCl Cl
7. Identify the major products P, Q and R in the following sequence of reactions: [NEET-2018]
+ CH CH CH Cl3 2 2
AnhydrousAlCl3 (i) O
2
(ii) H3O /+
P Q + R
P Q R
(1)
CH CH CH2 2 3
CHO
CH CH – OH3 2
, , (2)
CH CH CH2 2 3
CHO COOH
, ,
(3)
CH(CH )3 2
OH
CH – CO – CH3 3
,, (4) CH CH(OH)CH3 3
, ,
CH(CH )3 2
OH
133Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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Sol. Answer (3)
CH CH CH – 3 2 2
Cl + Al
Cl Cl
Cl
CH CH CH Cl AlCl3 2 2 3
+ –
(Incipient carbocation)
CH – CH – CH3 3
+
Cl
AlCl3
–
1, 2–H Shift
Now,
CH CH – CH3 3 –
CH – CH3
CH3
O2
HC – C – O – O – H 3
CH3
H /H O+
2
HydroperoxideRearrangement
OH
+CH – C – CH3 3
O
(R) (Q)
(P)
8. Which one is the most acidic compound? [NEET-2017]
(1)
OH
CH3
(2)
OH
(3)
OH
NO2
(4)
OH
NO2
NO2
O2N
Sol. Answer (4)
–NO2 group has very strong –I & –R effects.
9. The heating of phenyl-methyl ethers with HI produces. [NEET-2017]
(1) Ethyl chlorides (2) Iodobenzene
(3) Phenol (4) Benzene
Sol. Answer (3)
O – CH3
HI
OH
+ CH I3
134 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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10. The reaction,
OHNaH
OMe – I
Na
O
Me
can be classified as [NEET-2016]
(1) Williamson alcohol synthesis reaction (2) Williamson ether synthesis reaction
(3) Alcohol formation reaction (4) Dehydration reaction
Sol. Answer (2)
Fact
11. Which of the following reagents would distinguish cis-cyclopenta-1, 2-diol from the trans-isomer?
[NEET-2016]
(1) Aluminium isopropoxide (2) Acetone
(3) Ozone (4) MnO2
Sol. Answer (2)
cis-cylopenta-1, 2-diol can form cyclic ketal whereas tran-cyclopenta-1, 2-diol can't form cyclic ketal.
OH
OH
+ O = C
CH3
CH3
O
O
C
CH3
CH3
12. Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the
following functional group? [Re-AIPMT-2015]
(1) –CHCl2
(2) –CHO (3) –CH2Cl (4) –COOH
Sol. Answer (2)
OH OH
CHO(i) CHCl + NaOH
3
(ii) H+
13. Which of the following is not the product of dehydration of OH ? [Re-AIPMT-2015]
(1) (2) (3) (4)
Sol. Answer (4)
More stable carbocation can't be rearranged to a less stable carbocation.
135Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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14. Which of the following reaction(s) can be used for the preparation of alkyl halides?
(I) 2anh. ZnCl
3 2CH CH OH HCl (II) CH
3CH
2OH + HCl
(III) (CH3)3COH + HCl (IV) 2
anh. ZnCl
3 2(CH ) CHOH HCl
[Re-AIPMT-2015]
(1) (IV) only (2) (III) and (IV) only
(3) (I), (III) and (IV) only (4) (I) and (II) only
Sol. Answer (3)
The reactions of primary and secondary alcohols with HCl require the presence of a catalyst ZnCl2.
15. The reaction CH —C—ONa + CH CH Cl 3 3 2
CH3
CH3
—NaCl
CH —C—O—CH —CH3 2 3
CH3
CH3
is called [AIPMT-2015]
(1) Gatterman - Koch reaction (2) Williamson synthesis
(3) Williamson continuous etherification process (4) Etard reaction
Sol. Answer (2)
In Williamson synthesis, R – O Na + R – + X R – O – R + Na X
S 2 attactN
(ether)
As attack is SN2 R – X should be less sterically hindered i.e, R – X should be H
3C – X or 1° R – X
So,
S 2 attactN
H C C ONa + H C CH Cl3 3 2
+
CH3
CH3
H C C O CH CH3 2 3
CH3
CH3
16. Among the following sets of reactants which one produces anisole ? [AIPMT-2014]
(1) CH3CHO; RMgX (2) C
6H
5OH; NaOH; CH
3I
(3) C6H
5OH; netural FeCl
3(4) C
6H
5–CH
3; CH
3COCl; AlCl
3
Sol. Answer (2)
OHO Na
NaOH CH – I3
O
CH3
(Anisole)
17. Identity Z in the sequence of reactions CH3CH
2CH = CH
2 2 2HBr /H OY 2 5C H ONa Z [AIPMT-2014]
(1) CH3–(CH
2)3–O–CH
2CH
3
(2) (CH3)2CH
2–O–CH
2CH
3
(3) CH3(CH
2)4–O–CH
3
(4) CH3CH
2–CH(CH
3)–O–CH
2CH
3
136 Alcohols, Phenols and Ethers Solutions of Assignment (Level-II)
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Sol. Answer (1)
H C CH CH3 2
CH2
HBr / H O2 2
H C CH CH3 2
2
CH2
Br
(Y)
C H CH ONa3 2
H C CH3 2
CH CH2 2
O
H C2
H C3
Z is H3C – (CH
2)3
– O – CH2 – CH
3
18. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated H ?
[NEET-2013]
(1)
CH3
CH –CH –CH–O–CH3 2 3
(2)
CH3
CH –C–O–CH3 3
CH3
(3) CH –CH–CH –O–CH3 2 3
CH3
(4) CH3–CH
2–CH
2–CH
2–O–CH
3
Sol. Answer (2)
19. In the following reaction :
CH3
H3C–C–CH = CH
2
CH3
H2O/H
Major Product Minor ProductA B+
The major product is [AIPMT (Prelims)-2012]
(1)
CH3
H3C–C – CH–CH
3
CH3OH
(2)
CH3
H3C–C – CH –CH
2 2
CH3
OH
(3)
CH3
H3C–C – CH–CH
3
CH3OH
(4)
CH3
CH2–C–CH –CH
2 3
CH3OH
Sol. Answer (3)
CH = CH2
H+
CH – CH3
+
1 – 2 Ml Shift–
CH CH3
CH3
H O2
OH H
A
CH CH3
CH3
H O2
OH
CH3
H3C
CH3
CH CH3
137Solutions of Assignment (Level-II) Alcohols, Phenols and Ethers
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20. In the following reactions,
(a)
3
H /Heat3 3
Major Minorproduct product
CH|
CH – CH– CH– CH A B|
OH
(b)
HBr, dark
in absence of peroxide MinorMajorProductProduct
A C D
The major products (A) and (C) are respectively [AIPMT (Prelims)-2011]
(1)
3
2 2 3
CH|
CH C CH – CH and
3
3 2 3
CH|
CH C CH – CH|
Br
(2)
3
2 2 3
CH|
CH C CH – CH and
3
2 2 3
CH|
CH CH CH – CH
|
Br
(3)
3
3 3
CH|
CH C CH– CH and
3
3 2 3
CH|
CH C CH – CH|
Br
(4)
3
3 3
CH|
CH C CH– CH and
3
3 3
CH|
CH CH CH – CH|
Br
Sol. Answer (3)
As per the reaction sequence
Br
and will form
21. Which one of the following compounds has the most acidic nature ? [AIPMT (Prelims)-2010]
(1) CH OH2 (2) OH (3) OH (4)
OH
CH
Sol. Answer (2)
OH is the most acidic.
22. Which one is most reactive towards electrophilic reagent? [AIPMT (Prelims)-2010]
(1)
CH3
OH(2)
CH3
CH OH2
(3)
CH3
NHCOCH3 (4)
CH3
OCH3
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Sol. Answer (1)
Due to greater e– releasing effect.
23. Among the following four compounds
a. Phenol b. Methyl phenol c. m-Nitrophenol d. P-nitrophenol
The acidity order is [AIPMT (Mains)-2010]
(1) d > c > a > b (2) c > d > a > b (3) a > d > c > b (4) b > a > c > d
Sol. Answer (1)
Withdrawing group increasing the acidic character and electron donating group decreases the acidic characters.
24. When glycerol is treated with excess of HI, it produces [AIPMT (Mains)-2010]
(1) 2-iodopropane (2) Allyl iodide (3) Propene (4) Glycerol triiodide
Sol. Answer (1)
CH OH2—
CH OH—
CH OH2—
+ HIexcess
CH3
CH I—
CH3
25. Which one of the following compounds will be most readily dehydrated ? [AIPMT (Mains)-2010]
(1)OH
O
CH3
(2)OH
O
H C3(3)
CH3
OHO
(4)CH3
OH
O
Sol. Answer (3)
As carbocation intermediate, more the stability of carbocation, faster the rate of dehydration.
26. Which of the following conformers for ethylene glycol is most stable? [AIPMT (Mains)-2010]
(1)H
HH
H
OH
OH
(2)
HH
OH
H
OH
H(3)
HH
OH
H
H
HO(4)
HH
OH
H
H OH
Sol. Answer (4)
Intramolecular H-bonding.
27. Consider the following reaction,
Ethanol 3PBr X alc. KOH Y2 4
2
(i)H SO room temperature
(ii)H O, heatZ ; [AIPMT (Prelims)-2009]
(1) CH3CH
2–O–CH
2–CH
3(2) CH
3–CH
2–O–SO
3H (3) CH
3CH
2OH (4) CH
2 = CH
2
Sol. Answer (3)
OH
PBr3
Br
Alc KOH
CH = CH2 2
H SO , H O2 4 2Y
OH
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28. Consider the following reaction
Phenol 3 4
3
CH Cl Alkaline KMnO
Zn dust Anhydrous AlClX Y Z
The product Z is [AIPMT (Prelims)-2009]
(1) Benzaldehyde (2) Benzoic acid (3) Benzene (4) Toluene
Sol. Answer (2)
OHZn CH Cl
3
AlCl3
CH3
[ O ]COOH
X Y Z
29. H2COH.CH
2OH on heating with periodic acid gives [AIPMT (Prelims)-2009]
(1) 2HCOOH (2) CHO
CHO
(3) 2H
C = OH
(4) 2 CO2
Sol. Answer (3)
OH
OH
+ HIO .2H O4 2 2HCHO
30. The major organic product in the reaction, CH3 – O – CH(CH
3)
2 + H Product is [AIPMT (Prelims)-2006]
(1) CH3OH + (CH
3)2CHI (2) ICH
2OCH(CH
3)2
(3) CH O C(CH )3 3 2
I
(4) CH3 + (CH
3)2CHOH
Sol. Answer (4)
H C3
+ HI CH I +3
OHO
31. Ethylene oxide when treated with Grignard reagent yields [AIPMT (Prelims)-2006]
(1) Secondary alcohol (2) Tertiary alcohol
(3) Cyclopropyl alcohol (4) Primary alcohol
Sol. Answer (4)
32. Which one of the following compounds is most acidic? [AIPMT (Prelims)-2005]
(1) Cl–CH2–CH
2–OH (2)
NO2
OH
(3)OH
(4)OH
CH3
Sol. Answer (2)
OH
NO2
is the most acidic because of (–M & –I) effect of –NO2.
33. When 3, 3-dimethyl 2-butanol is heated with H2SO
4, the major product obtained is
(1) 2, 3-dimethyl 2-butene
(2) cis and trans isomers of 2, 3-dimethyl 2-butene
(3) 2, 3-dimethyl 1-butene
(4) 3, 3-dimethyl 1-butene
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Sol. Answer (1)
+ H+
+ 1 – 2 Ml–
Shift
+
OH
34. Decreasing order of reactivity of hydrogen halide acids in the conversion of ROH RX is
(1) HCl > HBr > HI > HF (2) HI > HBr > HCl > HF
(3) HF > HCl > HBr > HI (4) HF > HBr > HI > HCl
Sol. Answer (2)
R–OH + HX R – X
The HX reactivity order is HI > HBr > HCl > HF
35. More acidic than ethanol is
(1) CH3CH
2CH
2CH
2CH
2CH
2CH
2CH
3(2) CH
3CO
2CH
2CH
3
(3) CH3COCH
2COCH
3(4) CH
3COCH
3
Sol. Answer (3)
CH3
Active methyl more acidic than ethanol.
36. Which reagent converts propene to 1-propanol?
(1) H2O, H
2SO
4(2) B
2H
6, H
2O
2, OH–
(3) Hg(OAc)2, NaBH
4/H
2O (4) Aq. KOH
Sol. Answer (2)
OH?
The reagent must be B2H
6, H
2O
2/OH–
37. n-propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?
(1) PCl5
(2) Reduction
(3) Oxidation with potassium dichromate (4) Ozonolysis
Sol. Answer (3)
OH
[O]
OH
O
&
OH
[O]
O
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38. Which one of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?
(1) CH3CH(OH)CH
3(2) CH
3CH
2CH(OH)CH
3(3) CH
3OH (4) CH
3CH
2OH
Sol. Answer (3)
CH3OH + I
2 / OH– No reaction.
39. The general molecular formula, which represents the homologous series of alkanols is
(1) CnH
2n + 2O (2) C
nH
2nO
2(3) C
nH
2nO (4) C
nH
2n + 1O
Sol. Answer (1)
CnH
2n + 2O is alkanols.
40. On heating glycerol with conc. H2SO
4, a compound is obtained which has bad odour. The compound is
(1) Acrolein (2) Formic acid (3) Allyl alcohol (4) Glycerol sulphate
Sol. Answer (1)
CH OH2
CH OH + H SO2 2 4
CHO
CH OH2
(Conc.) (Acrolein)
41. The correct acidic order of the following is
OH OH OH
I. II. III.
CH3
NO2
(1) I > II > III (2) III > I > II (3) II > III > I (4) I > III > II
Sol. Answer (2)
OH
NO2
>
OH OH
CH3
>
42. When phenol is treated with CHCl3 and NaOH, the product formed is
(1) Benzaldehyde (2) Salicylaldehyde (3) Salicylic acid (4) Benzoic acid
Sol. Answer (2)
OH + CHCl + OH3
–OH
CHO
43. The compound which does not react with sodium, is
(1) CH3COOH (2) CH
3 – CHOH – CH
3(3) C
2H
5OH (4) CH
3 – O – CH
3
Sol. Answer (4)
O
CH3
CH3
+ Na No reaction
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44. Reaction of CH – CH2 2
O
with RMgX leads to formation of
(1) RCH2CH
2OH (2) RCHOHCH
3(3) RCHOHR (4) CHCH OH
2
R
R
Sol. Answer (1)
+ RMgX R
OH
45. Which of the following will not be soluble in sodium hydrogen carbonate?
(1) 2,4,6-trinitrophenol (2) Benzoic acid (3) o-Nitrophenol (4) Benzenesulphonic acid
Sol. Answer (3)
OH
ON
O
Intra H-bonding
Due to presence of intra H-bond, the ‘H’ is engaged, it becomes weak acid. Moreover NaHCO3 is a weaker
base. So, o–nitrophenol does not react with NaHCO3 and hence it is not soluble in NaHCO
3.
SECTION - C
Assertion-Reason Type Questions
1. A : p-nitrophenol has high pKa in comparison to o-nitrophenol.
R : In o-nitrophenol, intermolecular H-bonding is present.
Sol. Answer (4)
OH
NO2
>
O
H
N
O
O
Intra H–bond
Due to intra H bond in o-nitrophenol it is less acidic than p-nitrophenol.
So, Ka (p-nitrophenol) > K
a (o-nitrophenol)
So, pKa order p-nitrophenol < o-nitrophenol
2. A : When C2H
5–O–CH
3 is reacted with one mole of HI then C
2H
5OH & CH
3I is formed.
R : It is SN1 reaction.
Sol. Answer (3)
The -R gp 3° then only SN1 attack of I– takes place otherwise S
N2 attack takes place.
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So, in absence of 3°R, that –‘R’ is being attack by I– (or X–) which is less sterically crowded, because the
attack is SN2. Since S
N2 reactivity order follows as H
3C – I > 1°R – I > 2°R – I > 3°R – I
Therefore, H C CH3 2
O CH–
3
H+
HIH C CH
3 2OH
H C CH3 2
O CH3
H
I
S 2 attact
is easy and fastN
ICH3
3. A : When 3, 3-dimethyl butan-2-ol is heated in presence of concentrated H2SO
4 then 2, 3-dimethyl but-2-ene is
formed as major product.
R : In this reaction, carbocation is formed as an intermediate.
Sol. Answer (2)
H C3 C CH
CH3
CH3
Conc. H SO2 4CH
3
OH
H C3 C C
CH3
CH 3
CH3
+H and –H O+
2
H C3 C CH
CH3
CH3
CH 3
–CH shift3
H C3 C CH
CH3
CH3
CH 3
–H+
H C3 C C
CH3
CH3
CH 3
Note 2, 3-dimethyl
but-2-ene is formed because of –CH shift
takes place.
3
4. A : In esterification reaction, HCOOH is the most reactive acid among carboxylic acid.
R : Alcohol acts as nucleophile.
Sol. Answer (2)
As in esterification, steric crowding is dominant factor, so smaller is the ‘R’ group of carboxylic acid more is
(RCO2H) the reactivity.
is most acidic since ‘H’ is smallest.
5. A : Ethers can't be distilled upto dryness due to fear of explosion.
R : Due to the formation of superoxide, it is explosive.
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Sol. Answer (3)
Ethers absorb and react with oxygen from air, in presence of light, forming unstable peroxide that can deto-
nate with extreme violence when they become concentrated through evaporation or distillation and disturbed
by heat, shock or friction.
6. A : Phenol does not react with NaHCO3.
R : Phenol is less acidic than H2CO
3.
Sol. Answer (1)
Acidic strength order
OH
OH
So,
OH
+ NaHCO 3
O Na– +
+ H CO2 3
Reaction is not possible.
7. A : CH3
C COOH
O
gives haloform reaction.
R : It is more acidic than acetic acid.
Sol. Answer (2)
With I2 + NaOH and I
2 + Na
2CO
3, the iodoform (or haloform) test is given by pyruvic acid (H C
3C
O
CO H)2
.
H C3
C
O
C OH
O
I + NaOH2
CHI3
+ C C
O O
Na O–+
O–
Na+
+ H O2
+ NaI
But this reaction is not because that it is more acidic than acetic acid.
8. A : Diphenyl ether is prepared by Williamson synthesis.
R : This reaction generally proceed by SN1 mechanism.
Sol. Answer (4)
Diphenyl ether cannot be prepared by Williamsons's synthesis, because SN2 attack on ‘C’ of benzene is not
possible.
O Na+–
+
X O
Not possible
Wiiliamson's synthesis always follows SN2 attack.
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9. A : Grignard's reagent is prepared in the presence of ether.
R : Grignard's reagent is soluble and stable in ether.
Sol. Answer (1)
Grignard reagent is prepared in presence of ether solvent because Grignard reagent (RMgX) is more soluble
in ether solvent and stable as ether doesn't provid 'H+'.
etherR X Mg R MgX
10. A : CH3
C CH CH2
CH3
CH3
on hydroboration oxidation givesCH3
C CH CH3
CH3
OH CH3
as major product.
R : It involves the formation of carbocation so undergoes rearrangement.
Sol. Answer (4)
Hydroboration oxidation follows through formation of 4-membered cyclic Ts only and not via carbocation and
hence it gives anti markovnikov addition of H2O on alkene.
i.e., H C3 C CH
CH3
CH3
CH2
H C3 C CH
CH3
CH3
CH2
OH
11. A : Two moles of Grignard reagent is consumed in the formation of tertiary alcohol from ester followed by
hydrolysis.
R : One mole of Grignard reagent converts ester into Ketone and second mole of Grignard reagent adds to
Ketone.
Sol. Answer (1)
OH
R
R C R
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12. A : O
R
R
bond angle in ether is slightly greater than normal tetrahedral angle (109.5°).
R : The hybridisation of oxygen atom in ether is sp3.
Sol. Answer (2)
The B.A. in ether is slightly greater than normal tetrahedral angle 109.5°, this is because of repulsion between
two alkyl group.
13. A : CH3
C O CH3
CH3
CH3
on reaction with conc. HI gives CH3
C I
CH3
CH3
and CH3OH as major product.
R : This reaction proceed by SN1 mechanism.
Sol. Answer (1)
H C3 C O
CH3
CH3
CH3
H C3 C I + H COH
3
CH3
CH3
H+
H C3 C O
CH3
CH3
CH3
H
H C3 C
CH3
CH3
I
H C3 C
CH3
CH3
I
+ H COH3
In case of 3°R the attack of HI is SN1 and takes place at first.
14. A : Ortho-cresol is weaker acidic than meta-cresol.
R : It is due to ortho effect.
Sol. Answer (3)
Acidic order
OH
(o-cresol)
CH3
OH
(m-cresol)
CH3
This is because of +I effect.
Ortho effect is found in benzoic acid only.
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15. A : Among all ortho halophenol, fluorophenol is least acidic.
R : Ortho-fluorophenol forms intramolecular H-bond.
Sol. Answer (1)
F
Intra H–bond
H
O
Since only - ‘F’ can form H-bond among all halogen, therefore it is least acidic.
16. A : In esterification reaction alcohol act as nucleophile.
R : In this reaction O–H bond of alcohol is broken.
Sol. Answer (1)
In esterification reaction alcohol acts as nucleophile because there is lp present on OH of R – OH and since
O – H bond can be cleaved easily. So, it favours its nucleophility.
Act as Nu
17. A : Phenol is manufactured by Dow's process.
R : It involves the formation of benzyne intermediate.
Sol. Answer (2)
Dow’s process, involves the formation of benzyne intermediate but is not the correct reason.
Dow's process
Cl
1.NaOH
(P & T)
ONa
2. H O3
+
OH
18. A : Primary alcohol is prepared by the reaction of primary amine with HNO2.
R : Dimethyl amine is a primary amine but does not form methyl alcohol with HNO2.
Sol. Answer (3)
Primary alcohol is prepared by the reaction of primary amine with HNO2.
2 2 2 2R NH HNO R OH N H O
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19. A : The reactivity order of alcohols is 1° > 2° > 3° for the reaction in which O–H bond is broken.
R : The reactivity order of alcohol is 3° > 2° > 1° for the reaction in which C–O bond is broken.
Sol. Answer (2)
Reactivity order of alcohols for the reaction in which O – H bond is broken is in the order of
H3C – OH > 1°R – OH > 2°R – OH > 3°R – OH
As the stablity order of their conjugate base.
H3C – O– > 1°R – O– > 2°R – O– > 3°R – O– (because of +I effect of ‘R’ group)
Reactivity order of alcohol for the reaction in which C – O bond is broken is as
3°R – OH > 2°R – OH > 1°R – OH > H3C
– OH
Since stablity order of carbocation.
3°R+ > 2°R+ > 1°R+ > H3C+
20. A : The dehydration of ethyl alcohol in presence of Al2O
3 at 633 K gives ethene.
R : The reaction proceed through the formation of carbocation intermediate.
Sol. Answer (3)
H C CH3 2
OH
H C CH2 2
Al O2 3
633 K
H CH3 2C
OH
(I°R – OH) prefer to go E2 elimination reaction, so carbocation does not form in this reaction.
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